= Solution
{c}
Every vector decomposes as $v=Ev+(v-Ev)$, where $Ev\in\operatorname{im}E$ and $E(v-Ev)=0$. If $w$ lies in both the image and kernel, write $w=Eu$ to obtain $w=E^2u=Ew=0$. Thus
$$
V=\operatorname{im}E\oplus\ker E,
$$
and $E$ is the <projection> onto its <image> along its <kernel>. It is the identity on its image and zero on its kernel. Let $N=\dim V$ and $r=\operatorname{rank}E$. A basis adapted to this <direct sum> gives matrix $\operatorname{diag}(I_r,0_{N-r})$. Since $E$ is not the identity, $N-r>0$, and therefore
$$
\boxed{\det E=0.}
$$
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